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As @AngshukN has mentioned that both J-K flip-flops are negative edge-triggered and so, will be activated for falling edge.

Consider Output of $1^{st}$ flip-flop is $Q_A$ and Output of 2nd flip-flop is $Q_B$ and initially $Q_A=0,Q_B=0$ [Since, nothing is given in the question, we can assume like this and at the end, we can consider $(0,1),(1,0),(1,1)$ for both $Q_A$ and $Q_B$ and options can be eliminated easily] 

$1)$  Before $1^{st}$ falling edge of the clock: For first flip-flop, values of inputs J and K will not affect the output because flip-flop is not activated. So, Before the 1st falling edge, $Q_A=Q_1$ will remain same i.e. zero and since, output $Q_A$ is working as clock for the second flip-flop and there is no change of $1$ to $0$ for $Q_A$, it means second flip-flop is also not activated and so,output $Q_B$ will remain same as zero and so, before $1^{st}$ falling edge, $Q_2=\overline{Q_B}=1$

$2)$ At $1^{st}$ falling edge of the clock: For first flip-flop, $J=1,K=0$, it means, flip-flop will be in the set mode and make output $Q_A =Q_1=1$ and this $Q_A=Q_1=1$ will be same for next falling edge i.e. $2^{nd}$ falling edge. Now, for the $2^{nd}$ flip-flop, $Q_A$ works as clock and recently $Q_A$ changed from $0$ to $1$, not $1$ to $0$ which is requirement to activate $2^{nd}$ flip-flop and it means $Q_2=1$ as initially before the next falling edge of the clock i.e. second falling edge of the clock. So, after $1^{st}$ falling edge of the clock and before $2^{nd}$ falling edge of the clock, $Q_1=1$ and $Q_2=1$

$3)$ At $2^{nd}$ falling edge of the clock: For first flip-flop, $J=0,K=1$, it means this flip-flop is in reset mode and so, $Q_A=Q_1$ goes from $1$ to $0$ and will remain in the same state i.e $0$ before the $3^{rd}$ falling edge. Now, for the $2^{nd}$ flip-flop, since $Q_A$ is working as clock and recently $Q_A$ changed from $1$ to $0$ and so, $2^{nd}$ flip-flop will get activated and at this point of time, for this second flip-flop, $J=\overline{Q_B}= 1,K=1$($Q_B$ is taken from previous state of second flip-flop), so, second flip-flop will be in the toggle mode and so, $Q_B=1$ before the $3^{rd}$ falling edge of the clock and it means, $Q_2= \overline{Q_B}=0$ for before the $3^{rd}$ falling edge of the clock.

$4)$ At $3^{rd}$ falling edge of the clock: For first flip-flop, $J=1,K=0$, it means $Q_A=Q_1=1$ before the $4^{th}$ falling edge of the clock and since $Q_A$ changed from $0$ to $1$ at this point of time and so, second flip-flip will not be activated and $Q_2$ will remain in the same state as previous i.e. zero till $4^{th}$ falling edge of the clock.

$5)$ At $4^{th}$ falling edge of the clock: For first flip-flop, $J=1,K=1$, it means first flip-flop is in the toggle mode and so, $Q_A=Q_1$ changes from $1$ to $0$ and so, second flip-flop will be activated and at this point of time, for second flip-flop, $J= \overline{Q_B}= 0, K=1$ and so, second flip-flop is in reset mode and so, $Q_B=0$ till $5^{th}$ falling edge of the clock and so, $Q_2= \overline{Q_B}=1$ till $5^{th}$ falling edge of the clock.

$6)$ At $5^{th}$ falling edge of the clock: For first flip-flop, $J=1,K=0$, it means first flip-flop is in set mode and so, $Q_A=Q_1$ goes from $0$ to $1$ and will remain in the same state till $6^{th}$ falling edge. Since, $Q_A$ changed from $0$ to $1$, so, second flip-flop will not be activated and so, $Q_2$ will be in the same state i.e. $1$ till $6^{th}$ falling edge of the clock.  

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